3.26.50 \(\int \frac {-15 x^2+96 x^3+50 x^4+6 x^5+e^{-5+x} (10 x-67 x^2-64 x^3-15 x^4-x^5)}{-6+\log (4)} \, dx\)

Optimal. Leaf size=32 \[ \frac {x (5+x) \left (-e^{-5+x}+x\right ) \left (x-x^2 (5+x)\right )}{6-\log (4)} \]

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Rubi [B]  time = 0.26, antiderivative size = 124, normalized size of antiderivative = 3.88, number of steps used = 24, number of rules used = 4, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {12, 2196, 2176, 2194} \begin {gather*} -\frac {x^6}{6-\log (4)}+\frac {e^{x-5} x^5}{6-\log (4)}-\frac {10 x^5}{6-\log (4)}+\frac {10 e^{x-5} x^4}{6-\log (4)}-\frac {24 x^4}{6-\log (4)}+\frac {24 e^{x-5} x^3}{6-\log (4)}+\frac {5 x^3}{6-\log (4)}-\frac {5 e^{x-5} x^2}{6-\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*x^2 + 96*x^3 + 50*x^4 + 6*x^5 + E^(-5 + x)*(10*x - 67*x^2 - 64*x^3 - 15*x^4 - x^5))/(-6 + Log[4]),x]

[Out]

(-5*E^(-5 + x)*x^2)/(6 - Log[4]) + (5*x^3)/(6 - Log[4]) + (24*E^(-5 + x)*x^3)/(6 - Log[4]) - (24*x^4)/(6 - Log
[4]) + (10*E^(-5 + x)*x^4)/(6 - Log[4]) - (10*x^5)/(6 - Log[4]) + (E^(-5 + x)*x^5)/(6 - Log[4]) - x^6/(6 - Log
[4])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-15 x^2+96 x^3+50 x^4+6 x^5+e^{-5+x} \left (10 x-67 x^2-64 x^3-15 x^4-x^5\right )\right ) \, dx}{-6+\log (4)}\\ &=\frac {5 x^3}{6-\log (4)}-\frac {24 x^4}{6-\log (4)}-\frac {10 x^5}{6-\log (4)}-\frac {x^6}{6-\log (4)}+\frac {\int e^{-5+x} \left (10 x-67 x^2-64 x^3-15 x^4-x^5\right ) \, dx}{-6+\log (4)}\\ &=\frac {5 x^3}{6-\log (4)}-\frac {24 x^4}{6-\log (4)}-\frac {10 x^5}{6-\log (4)}-\frac {x^6}{6-\log (4)}+\frac {\int \left (10 e^{-5+x} x-67 e^{-5+x} x^2-64 e^{-5+x} x^3-15 e^{-5+x} x^4-e^{-5+x} x^5\right ) \, dx}{-6+\log (4)}\\ &=\frac {5 x^3}{6-\log (4)}-\frac {24 x^4}{6-\log (4)}-\frac {10 x^5}{6-\log (4)}-\frac {x^6}{6-\log (4)}-\frac {10 \int e^{-5+x} x \, dx}{6-\log (4)}+\frac {15 \int e^{-5+x} x^4 \, dx}{6-\log (4)}+\frac {64 \int e^{-5+x} x^3 \, dx}{6-\log (4)}+\frac {67 \int e^{-5+x} x^2 \, dx}{6-\log (4)}-\frac {\int e^{-5+x} x^5 \, dx}{-6+\log (4)}\\ &=-\frac {10 e^{-5+x} x}{6-\log (4)}+\frac {67 e^{-5+x} x^2}{6-\log (4)}+\frac {5 x^3}{6-\log (4)}+\frac {64 e^{-5+x} x^3}{6-\log (4)}-\frac {24 x^4}{6-\log (4)}+\frac {15 e^{-5+x} x^4}{6-\log (4)}-\frac {10 x^5}{6-\log (4)}+\frac {e^{-5+x} x^5}{6-\log (4)}-\frac {x^6}{6-\log (4)}-\frac {5 \int e^{-5+x} x^4 \, dx}{6-\log (4)}+\frac {10 \int e^{-5+x} \, dx}{6-\log (4)}-\frac {60 \int e^{-5+x} x^3 \, dx}{6-\log (4)}-\frac {134 \int e^{-5+x} x \, dx}{6-\log (4)}-\frac {192 \int e^{-5+x} x^2 \, dx}{6-\log (4)}\\ &=\frac {10 e^{-5+x}}{6-\log (4)}-\frac {144 e^{-5+x} x}{6-\log (4)}-\frac {125 e^{-5+x} x^2}{6-\log (4)}+\frac {5 x^3}{6-\log (4)}+\frac {4 e^{-5+x} x^3}{6-\log (4)}-\frac {24 x^4}{6-\log (4)}+\frac {10 e^{-5+x} x^4}{6-\log (4)}-\frac {10 x^5}{6-\log (4)}+\frac {e^{-5+x} x^5}{6-\log (4)}-\frac {x^6}{6-\log (4)}+\frac {20 \int e^{-5+x} x^3 \, dx}{6-\log (4)}+\frac {134 \int e^{-5+x} \, dx}{6-\log (4)}+\frac {180 \int e^{-5+x} x^2 \, dx}{6-\log (4)}+\frac {384 \int e^{-5+x} x \, dx}{6-\log (4)}\\ &=\frac {144 e^{-5+x}}{6-\log (4)}+\frac {240 e^{-5+x} x}{6-\log (4)}+\frac {55 e^{-5+x} x^2}{6-\log (4)}+\frac {5 x^3}{6-\log (4)}+\frac {24 e^{-5+x} x^3}{6-\log (4)}-\frac {24 x^4}{6-\log (4)}+\frac {10 e^{-5+x} x^4}{6-\log (4)}-\frac {10 x^5}{6-\log (4)}+\frac {e^{-5+x} x^5}{6-\log (4)}-\frac {x^6}{6-\log (4)}-\frac {60 \int e^{-5+x} x^2 \, dx}{6-\log (4)}-\frac {360 \int e^{-5+x} x \, dx}{6-\log (4)}-\frac {384 \int e^{-5+x} \, dx}{6-\log (4)}\\ &=-\frac {240 e^{-5+x}}{6-\log (4)}-\frac {120 e^{-5+x} x}{6-\log (4)}-\frac {5 e^{-5+x} x^2}{6-\log (4)}+\frac {5 x^3}{6-\log (4)}+\frac {24 e^{-5+x} x^3}{6-\log (4)}-\frac {24 x^4}{6-\log (4)}+\frac {10 e^{-5+x} x^4}{6-\log (4)}-\frac {10 x^5}{6-\log (4)}+\frac {e^{-5+x} x^5}{6-\log (4)}-\frac {x^6}{6-\log (4)}+\frac {120 \int e^{-5+x} x \, dx}{6-\log (4)}+\frac {360 \int e^{-5+x} \, dx}{6-\log (4)}\\ &=\frac {120 e^{-5+x}}{6-\log (4)}-\frac {5 e^{-5+x} x^2}{6-\log (4)}+\frac {5 x^3}{6-\log (4)}+\frac {24 e^{-5+x} x^3}{6-\log (4)}-\frac {24 x^4}{6-\log (4)}+\frac {10 e^{-5+x} x^4}{6-\log (4)}-\frac {10 x^5}{6-\log (4)}+\frac {e^{-5+x} x^5}{6-\log (4)}-\frac {x^6}{6-\log (4)}-\frac {120 \int e^{-5+x} \, dx}{6-\log (4)}\\ &=-\frac {5 e^{-5+x} x^2}{6-\log (4)}+\frac {5 x^3}{6-\log (4)}+\frac {24 e^{-5+x} x^3}{6-\log (4)}-\frac {24 x^4}{6-\log (4)}+\frac {10 e^{-5+x} x^4}{6-\log (4)}-\frac {10 x^5}{6-\log (4)}+\frac {e^{-5+x} x^5}{6-\log (4)}-\frac {x^6}{6-\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 37, normalized size = 1.16 \begin {gather*} \frac {x^2 \left (-e^x+e^5 x\right ) \left (-5+24 x+10 x^2+x^3\right )}{e^5 (-6+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*x^2 + 96*x^3 + 50*x^4 + 6*x^5 + E^(-5 + x)*(10*x - 67*x^2 - 64*x^3 - 15*x^4 - x^5))/(-6 + Log[4
]),x]

[Out]

(x^2*(-E^x + E^5*x)*(-5 + 24*x + 10*x^2 + x^3))/(E^5*(-6 + Log[4]))

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fricas [A]  time = 1.06, size = 52, normalized size = 1.62 \begin {gather*} \frac {x^{6} + 10 \, x^{5} + 24 \, x^{4} - 5 \, x^{3} - {\left (x^{5} + 10 \, x^{4} + 24 \, x^{3} - 5 \, x^{2}\right )} e^{\left (x - 5\right )}}{2 \, {\left (\log \relax (2) - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^5-15*x^4-64*x^3-67*x^2+10*x)*exp(x-5)+6*x^5+50*x^4+96*x^3-15*x^2)/(2*log(2)-6),x, algorithm="fr
icas")

[Out]

1/2*(x^6 + 10*x^5 + 24*x^4 - 5*x^3 - (x^5 + 10*x^4 + 24*x^3 - 5*x^2)*e^(x - 5))/(log(2) - 3)

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giac [A]  time = 0.35, size = 52, normalized size = 1.62 \begin {gather*} \frac {x^{6} + 10 \, x^{5} + 24 \, x^{4} - 5 \, x^{3} - {\left (x^{5} + 10 \, x^{4} + 24 \, x^{3} - 5 \, x^{2}\right )} e^{\left (x - 5\right )}}{2 \, {\left (\log \relax (2) - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^5-15*x^4-64*x^3-67*x^2+10*x)*exp(x-5)+6*x^5+50*x^4+96*x^3-15*x^2)/(2*log(2)-6),x, algorithm="gi
ac")

[Out]

1/2*(x^6 + 10*x^5 + 24*x^4 - 5*x^3 - (x^5 + 10*x^4 + 24*x^3 - 5*x^2)*e^(x - 5))/(log(2) - 3)

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maple [B]  time = 0.06, size = 87, normalized size = 2.72

method	result	size
derivativedivides	\(\frac {-9875 \,{\mathrm e}^{x -5} \left (x -5\right )-12250 \,{\mathrm e}^{x -5}-3105 \,{\mathrm e}^{x -5} \left (x -5\right )^{2}-474 \,{\mathrm e}^{x -5} \left (x -5\right )^{3}-35 \,{\mathrm e}^{x -5} \left (x -5\right )^{4}-{\mathrm e}^{x -5} \left (x -5\right )^{5}-5 x^{3}+24 x^{4}+10 x^{5}+x^{6}}{2 \ln \relax (2)-6}\)	\(87\)
default	\(\frac {-9875 \,{\mathrm e}^{x -5} \left (x -5\right )-12250 \,{\mathrm e}^{x -5}-3105 \,{\mathrm e}^{x -5} \left (x -5\right )^{2}-474 \,{\mathrm e}^{x -5} \left (x -5\right )^{3}-35 \,{\mathrm e}^{x -5} \left (x -5\right )^{4}-{\mathrm e}^{x -5} \left (x -5\right )^{5}-5 x^{3}+24 x^{4}+10 x^{5}+x^{6}}{2 \ln \relax (2)-6}\)	\(87\)
risch	\(\frac {x^{6}}{2 \ln \relax (2)-6}+\frac {10 x^{5}}{2 \ln \relax (2)-6}+\frac {24 x^{4}}{2 \ln \relax (2)-6}-\frac {5 x^{3}}{2 \ln \relax (2)-6}+\frac {\left (-x^{5}-10 x^{4}-24 x^{3}+5 x^{2}\right ) {\mathrm e}^{x -5}}{2 \ln \relax (2)-6}\)	\(87\)
norman	\(-\frac {5 x^{3}}{2 \left (\ln \relax (2)-3\right )}+\frac {12 x^{4}}{\ln \relax (2)-3}+\frac {5 x^{5}}{\ln \relax (2)-3}+\frac {x^{6}}{2 \ln \relax (2)-6}+\frac {5 x^{2} {\mathrm e}^{x -5}}{2 \left (\ln \relax (2)-3\right )}-\frac {12 x^{3} {\mathrm e}^{x -5}}{\ln \relax (2)-3}-\frac {5 x^{4} {\mathrm e}^{x -5}}{\ln \relax (2)-3}-\frac {x^{5} {\mathrm e}^{x -5}}{2 \left (\ln \relax (2)-3\right )}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^5-15*x^4-64*x^3-67*x^2+10*x)*exp(x-5)+6*x^5+50*x^4+96*x^3-15*x^2)/(2*ln(2)-6),x,method=_RETURNVERBOSE
)

[Out]

1/2/(ln(2)-3)*(-9875*exp(x-5)*(x-5)-12250*exp(x-5)-3105*exp(x-5)*(x-5)^2-474*exp(x-5)*(x-5)^3-35*exp(x-5)*(x-5
)^4-exp(x-5)*(x-5)^5-5*x^3+24*x^4+10*x^5+x^6)

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maxima [A]  time = 0.48, size = 52, normalized size = 1.62 \begin {gather*} \frac {x^{6} + 10 \, x^{5} + 24 \, x^{4} - 5 \, x^{3} - {\left (x^{5} + 10 \, x^{4} + 24 \, x^{3} - 5 \, x^{2}\right )} e^{\left (x - 5\right )}}{2 \, {\left (\log \relax (2) - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^5-15*x^4-64*x^3-67*x^2+10*x)*exp(x-5)+6*x^5+50*x^4+96*x^3-15*x^2)/(2*log(2)-6),x, algorithm="ma
xima")

[Out]

1/2*(x^6 + 10*x^5 + 24*x^4 - 5*x^3 - (x^5 + 10*x^4 + 24*x^3 - 5*x^2)*e^(x - 5))/(log(2) - 3)

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mupad [B]  time = 0.10, size = 62, normalized size = 1.94 \begin {gather*} \frac {5\,x^2\,{\mathrm {e}}^{x-5}-24\,x^3\,{\mathrm {e}}^{x-5}-10\,x^4\,{\mathrm {e}}^{x-5}-x^5\,{\mathrm {e}}^{x-5}-5\,x^3+24\,x^4+10\,x^5+x^6}{\ln \relax (4)-6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((96*x^3 - 15*x^2 - exp(x - 5)*(67*x^2 - 10*x + 64*x^3 + 15*x^4 + x^5) + 50*x^4 + 6*x^5)/(2*log(2) - 6),x)

[Out]

(5*x^2*exp(x - 5) - 24*x^3*exp(x - 5) - 10*x^4*exp(x - 5) - x^5*exp(x - 5) - 5*x^3 + 24*x^4 + 10*x^5 + x^6)/(l
og(4) - 6)

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sympy [B]  time = 0.17, size = 71, normalized size = 2.22 \begin {gather*} \frac {x^{6}}{-6 + 2 \log {\relax (2 )}} + \frac {5 x^{5}}{-3 + \log {\relax (2 )}} + \frac {12 x^{4}}{-3 + \log {\relax (2 )}} - \frac {5 x^{3}}{-6 + 2 \log {\relax (2 )}} + \frac {\left (- x^{5} - 10 x^{4} - 24 x^{3} + 5 x^{2}\right ) e^{x - 5}}{-6 + 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**5-15*x**4-64*x**3-67*x**2+10*x)*exp(x-5)+6*x**5+50*x**4+96*x**3-15*x**2)/(2*ln(2)-6),x)

[Out]

x**6/(-6 + 2*log(2)) + 5*x**5/(-3 + log(2)) + 12*x**4/(-3 + log(2)) - 5*x**3/(-6 + 2*log(2)) + (-x**5 - 10*x**
4 - 24*x**3 + 5*x**2)*exp(x - 5)/(-6 + 2*log(2))

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